Parabola

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Mathematics / Parabola

Parabola

In the course of the next three chapters, we�ll be studying conic sections, which as the name suggests, havesomething to do with sections of a cone. Different ways of cutting a cone using a plane will give rise to differentconics.

We consider a (three-dimensional) double right circular as shown in the figure below, extended to infinity on bothsides, and with a semi-vertical angle θ .

Vθ

A double right circular cone extended to infinity on bothsides and with a semi-vertical angle . denotes the vertex of this double-cone

θ V

Fig - 01

Now we use a plane to cut this double cone. Refer to the next figure and see that, depending on how we orient ourplane, we give rise to

(a) circles - we�ve already discussed circles in the previous chapter; we thus see thatcircles are a particular sort of conic sections with the intersecting plane atinclination / 2π .

(b) parabola - we�ll discuss parabolas in this chapter; the intersecting plane for a parabolais parallel to the slant of the cone, i.e., at an angle θ .

(c) ellipse - the intersecting plane is at an angle ( / 2α θ α π> ≠ since then a circlewill be formed)

(d) hyperbola - the intersecting plane is at an angle ;α θ< in this case, the planecuts both the top and bottom halves of the cone



θ

Fig - 02

parabola

α θ =

α θ >

α = 2π /circle

ellipse

θ

α θ <

hyperbola

Depending on the orientation of the intersecting plane, different types of conic section will be generated from the double cone

The next figure shows that even straight lines, pairs of straight lines and in fact even a point can be considered assections of a cone. This is because these can be formed if the intersecting plane passes through the vertex of thedouble cone.

A point is formed if the plane passes through only the vertex

A straight line is formed if the plane is tangent to the double-cone's surface

A pair of straight lines is formed if the plane passer through the vertex and intersects both the top and bottom halves of the cone

Fig - 03

These last three cases are termed degenerate cases of the conic sections. In our subsequent discussion, by conicsections we will always imply only the no-degenerate cases, i.e, circles, parabolas, ellipses or hyperbolas.



A very important property that a conic section C satisfies is this. It is the locus of a moving point P such that P'sdistance from a fixed point is always in a constant ratio to its (perpendicular) distance from a fixed line. The fixedpoint is called the focus of C while the fixed line is called the directrix of C. The constant ratio is called theeccentricity of C and is denoted by e.

P

F a, b( )

X

L= 0

The conic C

A conic section is the locus ofa moving point such that

where is the perpendiculardistance of from the

= 0 and is the (fixed) .

CP

PXP

L Fdirectrix

focus

Fig - 04

PFPX

= e a ( constant)

For a circle, the eccentricity e is 0 because while the fixed point (the focus) is the centre of the circle, the fixed lineis assumed to be at infinity. Thus, PX in the figure above always remains infinitely large so that e = 0.

For the remaining conic sections, we have

Parabola : e = 1Ellipse : e < 1Hyperbola : e > 1

Our discussions in this and the next two chapter deal separately with these three types of conics.

As just discussed, a parabola is the locus of a moving point P such that its distance from a fixed point (the focus F)is always equal to its distance from a fixed line L = 0, the directrix.

For a start (and to make things easier), we assume F to be the point (a, 0) and the line L to be 0.L x a≡ + = Theorigin then lies mid-way between F and L. There�s no loss of generality in doing so since howsoever F and L maylie in the plane, we can always (by a suitable choice of the axes) make the origin of our axes lie half-way betweenF and L and the y-axis parallel to L.

F a, ( 0)

(focus)O

L= 0(Directrix)

y

x

Fig - 05

We have chosen our co-ordinate axes in such a way so that the origin lies mid-way between and = 0 and the -axis is parallel to = 0.Thus, the co-ordinates of the focus will be of the form ( , 0) while the directrix = 0 will be given by + = 0

OF L y

LF

aL

x a

Observe that since the origin O is equidistant from F and L = 0, it too lies on the parabola. Thus, our parabola willpass through the origin.

Section - 1 BASIC EQUATIONS OF A PARABOLA



To find the equation representing the parabola, we assume the co-ordinates of point P lying on it as (x, y). Thus,2 2( )PF x a y= − +

and # ( )P L x a= +By # we mean the distanceof from the line 0

P LP L

=

Fig - 06

P x, y( )

F a, ( 0)

x + a = 0

y

x

Note that = ( � ) + PF x a y2 2

M

while # = = ( + )P L PM x a

By the definition of a parabola, these two distances must always be equal so that2 2 2( ) ( )x a y x a− + = +

2 4 : Equation of a parabolay ax⇒ =

This is referred to as the standard equation of the parabola with focus F(a, 0) and directrix 0L x a≡ + = .

When we actually plot the curve 2 4 ,y ax= we obtain the following shape. Any point on this curve is alwaysequidistant from F (a, 0) and 0L x a≡ + = .

Fig - 07

P

F a, ( 0)

No matter where may lie

on = 4 , its distancefrom ( , 0) is always equalto its distance from

Py ax

F a

2

L = x + a = 0

O

L = x + a = 0

The line y = 0 which passes through the focus F(a, 0) as is perpendicular to the directrix x + a = 0 will be termedthe axis of the parabola. The origin, which is halfway between the focus and the directrix, will be termed thevertex of the parabola.

It should now be obvious for you to deduce what the curve 2 4y ax= will look like if a < 0. In this case, the focusF(a, 0) lies on the negative x-axis while the directrix x + a = 0 is to the right of the origin:



The curve = 4when < 0

y axa

2

Fig - 8

x a + = 0

( , 0)a

We can also deduce what the curve 2 4x ay= will look like; just interchange the role of x and y. The focus anddirectrix will change accordingly as described in the figure below

a > 0

x

y

O (vertex)

(0, )(focus)

a

y + a = 0(directrix)

a < 0

Fig - 09

x

y

O (vertex)

(0, )(focus)

a

y + a = 0(directrix)

In fact, we can now generalise this discussion to a parabola with vertex at V (h, k) and the axis parallel to the x-axis or the y-axis. Note that in these parabolas, the focus lies at a distance of |a| from the vertex along the axis ofthe parabola :

a > 0

F

V h, k( )Equation: ( ) 4 ( ) x – h = a y – k2

y

xa < 0

F

V h, k( )

Equation: ( ) 4 ( ) x – h = a y – k2

y

x

a > 0

FV h, k( )

Equation: ( ) 4 ( ) y – k = a x – h2

y

x

a < 0

F V h, k( )

Equation: ( ) x – h( ) = 4 y – k a2

y

x

Fig - 10



Plot the parabola given by the equation 2 4 4 4 0y y x− + − = .

Solution: The given equation can be rearranged as

2( 2) 4( 2)y x− = − −

This represents a parabola with vertex V(2, 2) and opening towards the left (the fourth case abovein Fig - 10) because a = �1 (negative).

The focus will lie at a distance 1 unit to the left of (2, 2), i.e. ,at (1, 2). The directrix will lie 1 unit to theright of (2, 2), i.e. it will be x = 3.

The following figure shows this parabola:

Fig - 11

V(2,2)

x

y

The required parabola.Can you find the co-ordinates of and ?

A, B C F ,(1 2)

(focus)

B

A

x = 3(Directrix)

C

____________________________________________________________________________________

We now focus on some important terminology pertaining to parabolas. We will be using the parabola 2 4y ax= asillustration for this purpose; however, this discussion is general.

LATUS-RECTUM: This is a unique chord for a given parabola. It is the chord passing through the focus andperpendicular to the axis of the parabola.

xAB is the latus rectum for this parabola

Fig - 12

y

A

x +a = 0

y = ax2 4

F a, ( 0)

B

The length of the latus rectum can easily be evaluated. Substituting x = a in the equationfor the parabola gives two corresponding values for y,

Example – 1



i.e., 2y a= ± corresponding to A and B. Thus, the length AB is 4a.

2

Length of Latus-Rectum of4

4a

y ax⇒ =

=

From now on, we will abbreviate the latus-rectum as LR. The length of the LR can alsobe seen to be 4a this way. Since A lies on the parabola, its distance from F, i.e., AFmust equal its distance from x +a = 0, which is 2a by inspection. Thus, AB = 2AF = 4a.

FOCAL-CHORD: Any chord passing through the focus of the parabola will be its focal chord. The LR is aparticular focal chord.

FOCAL DISTANCE: This, as the name suggests, is the distance of any point on the parabola from its focus,and which will by definition, by equal to its distance from the directrix.

x

The focal distance of ( , ) is which equals (the distance of from the directrix)The numerical value of

can be seen to be | + |

P x y PFPM.

P

PM x a

⊥

Fig - 13

y

M

x +a = 0

F

P x, y( )

Thus,

2

Focal distance of ( , )on 4

P x yx a

y ax= +

=

We now finally turn our attention to the most general form for the equation of a parabola, i.e., suppose that we aregiven an arbitrary fixed point F(h, k) and a fixed line 0L ax by c≡ + + = as the focus and directrix of the parabola.What is its equation is such a case?

We use the definition for the parabola, i.e. any point P(x, y) lying on it must be equidistant from F and L.

Fig - 14

P x y( , )

F h k( , )

L = ax + by + c = 0

Any point ( , )lying on the parabolamust be equidistant from and

P x y

F L

M



Referring to the figure above, we have

2 2PM PF=

22 2

2 2

( ) ( ) ( )ax by c x h y ka b+ +⇒ = − + −

+...(1)

2 2 2 2 2 2 2 2a x b y c abxy acx bcy⇒ + + + + +

2 2 2 2 2 2( ){ 2 2 }a b x y h k hx ky= + + + + − −

2 2 2 2 2 2 2 22 (2 ( ) 2 ) (2 ( ) 2 )b x a y abxy h a b ac x k a b bc y⇒ + − − + + − + +

2 2 2 2 2( )( ) 0a b h k c+ + + − =

Whatever the coefficients maybe, we see that the equation of a parabola in general has the form2 22 2 2 0Ax Hxy By Gx Fy C+ + + + + = ...(2)

So that (2) can be expressed in the form (1) (only then can (2) represent a parabola), it can be shown that thecoefficients in (2) must satisfy the relations

0A H GH B FG F C

∆ = ≠ and 2H AB=

This is left to the rigor-bent reader as an exercise

Find the equation of the parabola with focus F(�1, �2) and directrix 2 3 0.L x y≡ − + =

Solution: Assuming P(x, y) to be any point on the parabola, we must have, by virtue of P being equidistant fromF and L,

22 2( 2 3) ( 1) ( 2)

5x y x y− + = + + +

2 2 2 24 9 4 6 12 5 5 10 20 25x y xy x y x y x y⇒ + + − + − = + + + +

2 24 4 4 32 16 0x y xy x y⇒ + + + + + =

This is the required equation.

Find the length of the LR of the parabola 24 12 12 39 0y x y+ − + =

Example – 2

Example – 3



Solution: The equation can be rearranged as (verify)23 53

2 2y x − = − +

23 3 542 4 2

y x ⇒ − = − × × + This is of the form

2 4Y aX= −34

a =

i.e. this is a parabola with vertex 3 5,2 2

and axis parallel to the x-axis. The parabola opens to the

left. We are only concerned with the length of the LR which is simply 4a, i.e. 3 units.

The axis of a parabola is the line 1 : 3 4 4 0L x y+ − = and the tangent to it at the vertex is 2 : 4 3 7 0.L x y− + =The LR is 4 units in length. Find the equation of the parabola.

Solution: Consider for a moment, the co-ordinate axes system formed by L1 and L2.

Fig - 15

L1

P x, y( )B

AV

L2

L x y1 : 3 + 4 � 4 = 0L x y2 : 4 � 3 + 7 = 0

If L1 and L2 were truly the actual co-ordinate axes, the equation of the parabola would have been

2 4y ax=2 (Length of )PA LR PB⇒ = × ...(1)

Now, even if we use some other co-ordinate system (here the L1� L2 system), the relation (1) will stillhold since that only depends on lengths which are invariant with respect to the co-ordinate systemchosen. Thus, simply applying (1) will give us the required equation :

23 4 4 4 3 7

(4)5 5

x y x y + − − += ×

2(3 4 4) 20(4 3 7)x y x y⇒ + − = ± − +

Thus, as might have been expected, we�ll obtain two parabolas with the given property, one openingto the �left� and one to the �right� in the L1� L2 co-ordinate system.

Example – 4



Show that the locus of the middle points of all chords of the parabola 2 4y ax= which pass through the origin, isanother parabola.

Solution: Let OP be any such chord of the parabola, where O is the origin (0, 0) and P is the point (h, k) lyingsomewhere on the parabola. We then have

k2 = 4ah (∵ P lies on the parabola)

P' ( , )

Fig - 16

O

h2

k2

y

x

P h k( , )

The mid-point of OP, say P' is clearly ,2 2h k

.

Since 2 4 ,k ah= we have2

22 2k ha = 2 2y ax⇒ = ...(1)

where we used (x, y) instead of ,2 2h k

to specify the equation of the locus of P' in the conventionalx � y form.

(1) shows that the required locus is another parabola, which has the same vertex and axis. ____________________________________________________________________________________

PARAMETRIC FORM : The parabola 2 4y ax= is a lot of times specified not in the standard x y− form but

OF =2 4y ax instead in a parametric form, i.e., in terms of a parameter, say t.

The equation 2 4y ax= can be equivalently written in parametric form

2 , 2x at y at= =

This is easily verifiable by substitution. Thus, for any value of t, the point (at2, 2at) willalways lie on the parabola 2 4 .y ax= Different values of t give rise to different points onthe parabola.

The point (at2, 2at) is many times referred to as simply the point t.

Example – 5



From a point t on 2 4 ,y ax= a focal chord is drawn. Find the other end-point of this chord.

Solution: Let the other end-point be the point t1. Thus, the co-ordinates of the two ends of the chord are(at2, 2at) and 2

1 1( , 2 )at at .

Fig - 17

y

x

t

F a, ( 0)

t1

y = ax2 4

The equation of this chord is therefore

12 2 2

1

2 2 2y at at atx at at at

− −=− − ..(1)

Since this chord is a focal chord, the coordinates of F, i.e. (a, 0) must satisfy (1)

12 2 2

1

2 2 ( )( )

at a t ta at a t t− −⇒ =− −

21

11

tt t t

−⇒ =− +

2 21 1t tt t⇒ − − = −

1 1tt⇒ = −

This very widely used result simply states that the product of the two end-points t and t1 of any focalchord will be �1.

Let O be the origin and AB be any focal chord of the parabola 2 4 .y ax= A is the point t. Find

(a) the minimum area of OAB∆

(b) the locus of the centroid of OAB∆

Example – 6

Example – 7



Solution: Since AB is a focal chord, the point B is 1t

− . Thus, the co-ordinates of A and B are (at2, 2at)

and 2

2,a at t

− respectively.

(a) ( )

22 0

1 2area 2 02

1 1 1

aatt

aOAB att

∆ = −

(Expanding along C3)2

21 222

aa tt

= − −

2 1a tt

= − +

From calculus, we can easily deduce that the magnitude of 1tt

+ is minimum when 1t = ± the

minimum magnitude is hence 2: Thus, the minimum area of OAB∆ is 2a2. This occurs when ABis the LR, as might have been expected by symmetry.

Fig - 18

O

y

x

A

B

( 0)a,

Area of isminimum when

is the

∆OAB

AB LR

(b) Let (h, k) be the co-ordinates of the centroid of OAB∆ .

We have,

22 0

,3

aatth

+ +=

22 0

3

aattk

− +=

22

1 3 ,htt a

⇒ + = 1 3

2kt

t a− =



Squaring the second relation, we obtain

22

2 2

1 924kt

t a+ − =

Using the first relation in this,

2

2

3 924

h ka a

− =

2 212 8 9ah a k⇒ − =

224 33aa h k ⇒ − =

We now use (x, y) instead of ( h, k); the locus of the centroid is

2 23 43ay a x = −

____________________________________________________________________________________

LENGTH OF ANY: Through a point t, a focal chord is drawn in the parabola 2 4y ax= . The other

end-point of this chord is, as described earlier, 1.t

−

Therefore, the length of this chord is

2 22

2

22ta al at att t

= − + +

21 1 4a t t

t t = + − +

21a t

t = +

The minimum length for any focal chord is evidently obtained when 1,t = ± which givesus the LR. Thus, the smallest focal chord in any parabola is its LR.

Prove that the circle described on any focal chord of a parabola as diameter will touch the directrix.

FOCAL CHORD

Example – 8



Solution: Let the equation of the parabola be 2 4y ax= so that the equation of the circle described on any focalchord as diameter can be written as

22

2( ) ( 2 ) 0a ax at x y at yt t

− − + − + = ...(1)

The directrix is x + a = 0. To show that it touches the circle given by (1), it is sufficient to show thatsubstituting x = �a in (1) will give us only one value of y:

22

2( ) ( 2 ) 0a aa at a y at yt t

⇒ − − − − + − + = 2 2

2 2 22

( 1) 2 2 4 0t aa y at y at t+ ⇒ + + − − =

22 21 12 0y a t y a t

t t ⇒ + − + − =

This can evidently be written as a perfect square :2

1 0y a tt

− − =

1y a tt

⇒ = −

Only one value of y implies that the directrix touches the circle.

We can also prove the stated assertion geometrically simply by using the properties of a parabola.Consider any focal chord of the parabola, say AB; the next figures show the circle described on ABand the directrix of the parabola on different sketches.

Fig - 19

x

yA

O

B

Cx

yA

O

B

D

X

Y

F

Let O be the centre of the circle so described. To show that it touches the directrix, we have to showits radius, say OC, is equal to the its perpendicular distance of O from the directrix.



Consider the second sketch. In trapezium XABY, since O is the mid point of AB, we have

AX + BY = 2OD ...(1)

Since A lies on the parabola, we have by definition, AX = AF. Similarly, BY = BF. Thus,

AX + BY = AF + BF = AB = 2AO ...(2)

(1) and (2) give OD = OA which implies OD is a radius of the circle, i.e. OD = OC. Thus, the circletouches the directrix.

Through the vertex O of a parabola 2 4 ,y x= chords OP and OQ are drawn at right angles to one another. Showthat for all positions of P, PQ cuts the axis of the parabola at a fixed point. Also find the locus of the middle pointof PQ.

Solution: Let P and Q be the points t1 and t2 so that their co-ordinates are 21 1( , 2 )t t and 2

2 2( , 2 )t t respectively.

Fig - 20

O

y

x

P t , t( 2 )21 1

Q t , t( 2 )22 2

Since ,PO OQ⊥ we have

1 22 21 2

2 0 2 0 10 0

t tt t

− −× = −− −

2 21 2 1 24 0t t t t⇒ + =

1 2 1 2( 4) 0t t t t⇒ + =

1 2 4t t⇒ = − (since both t1 and t2 are non-zero)

The equation of the chord PQ is

1 1 22 2 21 1 2

2 2 2: y t t tPQx t t t− −=− −

21 2 1 1( )( 2 ) 2( )t t y t x t⇒ + − = −

2 21 2 1 1 2 1( ) 2 2 2 2t t y t t t x t⇒ + − − = −

1 2 1 22 ( ) 2 0x t t y t t⇒ − + + =

1 22 ( ) 8 0x t t y⇒ − + − = (since t1t2 = �4)

Example – 9



It is evident that this line always passes through (4, 0), whatever the value of t1 (and hence t2) may be.

The mid-point M(h, k) of PQ is given by

2 21 2 1 22 2,

2 2t t t th k+ += =

t1 and t2 can easily be eliminated from these relations using t1t2 = �4, to obtain

2 2 8k h= −

The locus of M is therefore

2 2( 4)y x= −

Let PQ be a variable focal chord of length l of the parabola 2 4 .y ax= From the vertex of this parabola, a

perpendicular is dropped into this focal chord; d is the length of this perpendicular. Prove that 2 34 .ld a=

Solution Let P be the point t so that Q is 1,t

− since PQ is the focal chord. We have

22

2

22a aPQ l at att t

= = − + +

21a t

t = +

(as in Example -7) ...(1)

The equation of PQ is

2

2

22

ayy at tax at xt

+− =− −

1 2 2y t x at

⇒ − = − ...(2)

The length of the perpendicular form the vertex, which is (0, 0), to PQ is,

2

2

14

ad

tt

= + −

22

24

1ad

tt

⇒ = +

...(3)

From (1) and (3), we have 2 34 .ld a=

Example – 10



1 1( , )P x y is a fixed point in the plane. How can we determine whether P lies inside or outside the parabola2 4 ?y ax=

Solution: Let us consider the case when a > 0. Then the parabola will open towards the right.

Consider the point 1 1( , )P x y lying in the plane as shown below :

Fig - 21

y

x

P x , y( )1 1

Q x , y( )1 0

y = ax2 4

Since Q lies on the parabola, 20 14y ax=

Since 1 0 ,y y> we have 21 14y ax> or 2

1 14 0y ax− > ; similarly, if P were to lie below the parabola,the same condition would be satisfied.

If P were to lie in the left half-plane, this condition would automatically be satisfied since x would benegative. Thus, we can now state the relative position of the point P(x1, y1) in terms of the equation

2( , ) 4S x y y ax= − as follows :

1 1( , ) 0S x y P> ⇒ lies outside the parabola

1 1( , ) 0S x y P= ⇒ lies on the parabola

1 1( , ) 0S x y P< ⇒ lies inside the parabola.

Verify that these conditions would hold true if a < 0 too.

A circle on any focal chord of a parabola as diameter cuts the curve again in P and Q. Show that PQ passesthrough a fixed point.

Solution: In example -8, we wrote the equation of a circle described on any focal chord as diameter in terms ofone of its end points t0.

20 02

0 0

2( ) ( 2 ) 0a ax at x y at yt t

− − + − + =

...(1)

Example – 11

Example – 12



The intersection of this circle with the parabola can be evaluated by using the parametric form of theparabola :

2 , 2x at y at= = ...(2)

Using (2) in (1), we obtain

2 2 2 2 20 02

0 0

1 1( ) 4 ( ) 0a t t t a t t tt t

− − + − + =

4 2 20 02

0 0

1 14 4 3 0t t t t tt t

⇒ + − − − − − =

...(3)

This has four roots in t, meaning that the circle intersects the parabola in four points, two of which are

obviously t0 and 0

1t

− (since these are the points using which the circle has been described in the first

place !).

Let the other two point P and Q be t1 and t2. Thus 0 1 20

1, , ,t t tt

− are the roots of (3):

0 1 2 0 1 20 0

1 10 ; 3t t t t t tt t

+ − + + = ⋅ − ⋅ ⋅ = −

1 2 00

1t t tt

⇒ + = − and 1 2 3t t = ...(4)

The co-ordinates of P and Q are 21 1( , 2 )at at and 2

2 2( , 2 ).at at Thus, the equation of PQ becomes :

1 1 22 2 21 1 2

2 2 ( ):( )

y at a t tPQx at a t t− −=− −

1 2 1 2( ) 2 2t t y x at t⇒ + = + ...(5)

Using (4) in (5), we obtain

00

1 2 6t y x at

− = +

This is always satisfied by (�3a, 0), no matter what the value of t0 may be. Thus, PQ always passesthrough the fixed point (�3a, 0).

Find the length of the intercept that the parabola 2 4y ax= makes on the line .y mx c= +

Example – 13



Solution: Let the end-points of the intercept be A and B :

Fig - 22

y

x

A

B

y = ax2 4

y=mx + c

The co-ordinates of A and B can be evaluated by simultaneously solving the equation of the line andthe parabola :

y mx c= +2 4y ax=

2( ) 4mx c ax⇒ + =2 2 2(2 4 ) 0m x mc a x c⇒ + − + = ...(1)

The first point to be noted from this quadratic is that the line will intersect the parabola in two distinctpoints only if the D of (1) is positive, because only then will two distinct values of x be obtained (thecase when m = ∞ gives two distinct points of intersection for one value of x, but that can be consideredseparately; in that case, the quadratic (1) will not be formed). We thus have,

D > 02 2 2( 2 )mc a m c⇒ − >

24 4amc a⇒ <

mc a⇒ < (a is +ve here)

Thus, the line will intersect the parabola only if a > mc. If a = mc, the line will be a tangent to theparabola. If a < cm, the line will not intersect the circle at all.

Assuming that D > 0, suppose that the roots of (1) are x1 and x2:

2

2 1 22 2

4 2 ,a mc cx x x xm m−+ = =

2 21 2 1 2 1 2( ) ( ) 4x x x x x x⇒ − = + −

2 2

4 2

(4 2 ) 4a mc cm m−= −

2 2 2 2 2

4

16 4 16 4a m c amc m cm

+ − −=

4

16 ( )a a mcm

−=



The corresponding y co-ordinates for x1 and x2 are

1 1y mx c= + and 2 2y mx c= +

so that 2 1 2 1( )y y m x x− = − .The length AB (intercept) is now simply

2 21 2 1 2( ) ( )AB x x y y= − + −

2 2 21 2 1 2( ) ( )x x m x x= − + −

22

4 ( )1

a a mcm

m−

= ⋅ +

This result need not be remembered. What is important is that you understand the underlying approach.

From this example, we can deduce one more useful thing : for a non-zero variable ,m ∈ " the lineay mxm

= + will always be a tangent to the circle. Verify that the point of contact of this tangent with

the parabola is 2

2, .a am m

We�ll discuss tangents in more detail in the next section.

Let P be the point (2a, 0) and QR be a variable chord of the parabola 2 4y ax= passing through P. Prove that

2 2

1 1 constantPQ PR

+ =

Solution: Since distances are involved, we can use the polar form for a straight line to write the equation of QR.Let PQ = r1 and PR = r2. Also, let QR be at an inclination .θ

Fig - 23

y

x

Q

y = ax2 4

P a, (2 0)

r1

r2

θ

R

Example – 14



The co-ordinates of any point on QR can be written as (in terms of its distance r from P)(2 cos , sin ).a r r+ θ θ If this lies on the parabola, we have

2 2sin 4 (2 cos )r a a rθ = + θ

2 2 2sin 4 cos 8 0r ar a⇒ θ − θ − = ...(1)

Since Q and R lie on the parabola, r1 and r2 must satisfy (1).Thus,

2

1 2 1 22 2

4 cos 8,sin sina ar r r rθ −+ = =

θ θ...(2)

Finally,

2 2 2 21 2

1 1 1 1PQ PR r r

+ = +

2 2

1 22

1 2( )r rr r+=

21 2 1 2

21 2

( ) 2( )

r r r rr r

+ −=

2 2 2

4 2

4

4

16 cos 16sin sin

64sin

a a

a

θ +θ θ=

θ(using (2))

2

14a

=

which is a constant.

Find all the points on the x-axis from which exactly three distinct chords of the circle 2 2 2x y a+ = can be drawn

which are bisected by the parabola 2 4 ( 0).y ax a= >

Solution: Let such a point be P(h, 0). We need to find the possible values that h can take.

Fig - 24

y

x

y = ax2 4

O

BC

X

A

P h, ( 0)We need to find all points such that three chords of the circle can be drawn passing through which are bisected by the parabola

P

P

Example – 15



Note that one such chord will always be simply along the x-axis because it is bisected by the parabolaat the origin.

Referring to Fig - 24, let C be the point t so that its co-ordinates are 2( , 2 ).at at We can write the

equation of the chord of the circle 2 2 2x y a+ = bisected at C as

2 2( , 2 ) ( , 2 )T at at S at at=

32 4tx y at at⇒ + = +

Since this passes through P(h, 0), we have

3 4th at at= +

2( (4 )) 0t at a h⇒ + − = ...(1)

One of the roots of (1) is t = 0 which corresponds to the case already mentioned, the diameter alongthe x-axis.

The other two roots are real and distinct from zero if

4 0ha

− >

4h a⇒ > ...(2)

If you think carefully, you will realise that an additional constraint has to be imposed, namely, a limit onthe value of t so that 2( , 2 )at at lies inside the circle, since only then will a chord be formed. Thus,

2 2 2 2( ) (2 )at at a+ <

2 5 2t⇒ < − ...(3)

From (1), we can see that (3) can equivalently be written as

4 5 2ha

− < −

( 5 2)h a⇒ < +

From (2) and (3), we see that the possible range of h is

( )( )4 , 5 2h a a∈ +



Q. 1 Find the equation of the parabola with vertex (2, �3) and focus (0, 5).

Q. 2 Find the equation of the parabola with vertex at (2,1) and directrix x = y �1.

Q. 3 Find the locus of the middle points of all chords of the parabola 2 4y ax= which are drawn throughthe vertex.

Q. 4 Find the length of the side of the equilateral triangle inscribed in the parabola 2 4y ax= , with onevertex at the origin.

Q. 5 Find the locus of a point which divides a chord of slope 2 of the parabola 2 4y x= internally in theratio 1 : 2.

Q. 6 Find the locus of the centre of the circle described on any focal chord of the parabola 2 4y ax= asdiameter.

Q. 7 Find the locus of the mid-points of all chords of length C of the parabola 2 4y ax= .

Q. 8 Two chords of the parabola 2 4y ax= passing through its vertex are perpendicular to each other. If p1

and p2 be the lengths of these chords, prove that

4 2 2/3 2 /3 31 2 1 2( ) (16 ( ))p p a p p= +

Q. 9 Consider a variable chord PQ through the focus of 2 4y ax= . The straight line joining P to the vertexcuts the line joining Q to the point (�a, 0) at R. Find the locus of R.

Q. 10 From the vertex of the parabola 2 4y ax= , a pair of chords is drawn, perpendicular to each other.With those chords as adjacent sides a rectangle is completed. Find the locus of the vertex of therectangle opposite to the vertex of the parabola.

TRY YOURSELF - I



In this section, we discuss tangents to parabolas and various properties that they will satisfy. As always, 2 4y ax=is the parabola used for illustration but the discussion can obviously be generalised.

TANGENT AT: Suppose a point P(x1 y1) is given on the parabola, and we need to find the equation of P (x1, y1) the tangent at P. The slope of the tangent can easily be evaluated by differentiating the

equation of the parabola :2 4y ax=

2 4dyy adx

⇒ =

2dy adx y

⇒ =

1 1at ( , ) 1

2P x y

dy adx y

⇒ =

Thus, the equation of the tangent is

1 11

2 ( )ay y x xy

− = −

21 1 12 2yy ax ax y⇒ = − +

21 1 12 2 4yy ax ax ax⇒ = − + 2

1 1( 4 )y ax=∵

1 12 ( )yy a x x⇒ = + ...(1)

This equation is sometimes written concisely as 1 1( , ) 0T x y = where the interpretationof T is understood to be as in (1)

TANGENT AT: A lot many times, the point P is specified in its parametric form and we need to find theP(at2, 2at) equation of the tangent in terms of the parameter t. This can be done by replacing

(x1, y1) with (at2, 2at) in the equation of the tangent just obtained :

1 12 ( )yy a x x= +

22 2 ( )aty a x at⇒ = +

2ty x at⇒ = +

This form of the tangent is very widely used and it is best for you to commit it tomemory.

Section - 2 TANGENTS



TANGENT OF Suppose that we are required to find the equation of the tangent with slope m.SLOPE m: We did so in the last section when we found the intersection of the line y mx c= + with

the parabola and obtained the condition for tangency to beacm

=

The equation of the tangent thus obtained was

ay mxm

= +

This line is a tangent to the parabola for all non-zero values of m.

The same equation can be obtained using a derivatives approach also. You are urged todo so as an exercise.

Show that the tangents at the extremities of any focal chord of a parabola intersect at right angles at the directrix.

Solution: Let the extremities of the focal chord be t1 and t2 so that t1t2 = �1. The equations to the tangents at t1and t2 are

21 1t y x at= +

22 2t y x at= +

The intersection point can easily be evaluated by solving these two equations to be1 2 1 2 1 2( , ( )) ( , ( ))at t a t t a a t t+ = − + since 1 2 1.t t = − Since the x-coordinate of the point of intersection

is �a, it lies on the directrix 0.x a+ = Also, note that the slopes of the two tangents are 11

1mt

= and

22

1mt

= so that 1 21 2

1 1.m mt t

= = − Thus, the tangents intersect at right angles on the directrix.

Fig - 25

y

xF a, ( 0)

Q at , at ( 2 )22 2

TP at , at ( 2 )2

1 1

The tangents at and intersect at right angleson the directrix, i.e

= 90º

P Q

PTQ∠

Note that T will also be the point at which the circle drawn on PQ as diameter touches the directrix,since PTQ∠ is 90° which then becomes an angle in the semi circle.

Example – 16



POINT OF : There�s a particular relation we obtained while solving the last example which is importantenough to be highlighted explicitly. If tangents are drawn to 2 4y ax= at t1 and t2, theirpoint of intersection will be

1 2 1 2, ( )at t a t t+

Prove that the orthocentre of any triangle formed by three tangents to a parabola lies on its directrix.

Solution: Let the equation of the parabola be 2 4y ax= so that its directrix is x + a = 0. We need to show thatthe x-coordinate of the orthocentre is �a.

Assume any three points on the parabola as t1, t2, t3. The points of intersections of the three tangentsto the parabola on these three points will be

1 2 1 2 2 3 2 3 3 1 3 1( , ( )), ( , ( )), ( , ( ))P at t a t t Q at t a t t R at t a t t+ + +

Let us find the equation of the altitude through P on QR.

The slope of QR is 2 3 3 1

2 3 3 1 3

( ) ( ) 1a t t a t tat t at t t+ − + =

−

Therefore, the equation of the altitude through P is

1 2 3 1 2( ) ( )y a t t t x at t− + = − −

3 1 2 1 2 3( )y t x a t t t t t⇒ + = + + ...(1)

By symmetry we can write the altitude from Q onto PR as

1 2 3 1 2 3( )y t x a t t t t t+ = + + ...(2)

By (1) � (2), we have

3 1 1 3( ) ( )t t x a t t− = −

x a⇒ = −

Thus, the stated assertion is true.

INTERSECTION

OF TANGENTS AT

t1 and t2.

Example – 17



Let P be any point on the parabola with focus F. A perpendicular PM is dropped from P onto the directrix asshown :

Fig - 26

y

x

R

F

PXM

S

Q N O

Prove that the tangent at P to the parabola bisects the angle FPM.

Solution: Let the equation of the parabola be 2 4 ,y ax= and let P be the point t. Thus, F is the point (a, 0) whileN is (�a, 0). By definition, we have

PF = PM

But PM is PX + XM i.e., 2 .at a+ Thus PF = a + at2.

Now, the equation of the tangent at P is

2ty x at= +

This intersects the x-axis at the point 2( ,0).Q at−

Thus,

FQ FO OQ= +

2a at= +

We see that in ,PFQ∆

FP = FQ

FPQ FQP⇒ ∠ = ∠ ...(1)

But since PM is parallel to the x -axis, we also have

FQP QPM∠ = ∠ ...(2)

From (1) and (2), we obtain FPQ QPM∠ = ∠ which means that the tangent at P bisects .FPM∠

Example – 18



Refer to Fig - 26 of the previous example. Prove that angle PFS is a right angle.

Solution: The equation of the tangent at 2( , 2 )P at at is

2ty x at= +

This intersects the directrix at a point given by 2( 1), a tat

−−

The slope of PF is

2 2

2 21PF

at tmat a t

= =− −

The slope of SF is

2

2( 1)

( 1)2 2SF

a ta ttm

a t

−−= =

− −

Since 1,PF SFm m× = − angle PFS is a right angle.

Refer to Fig - 26 of Example 18 again. Show that angle FRP is a right angle.

Solution: Once again, we use the equation of the tangent at 2( , 2 )P at at :

2ty x at= +

This intersects the y-axis at the point R(0, at). The slope of PR is simply the slope of the tangent at P,

i.e. 1

PRmt

=

The slope of RF is

00RFatm t

a−= = −−

Since 1,PR RFm m× = − angle FRP is a right angle.

Example – 19

Example – 20



The result of the last three examples are important, so we summarize them here :

* (A) The tangent at any point on a parabola bisects the angle between the focal chord through thatpoint and the perpendicular on the directrix from that point (Example-18).

* (B) The portion of the tangent to a parabola cut-off between the directrix and the curve subtends aright angle at the focus (Example -19).

* (C) The perpendicular dropped from the focus onto any tangent to a parabola is concurrent with thattangent and the tangent at the vertex (Example -20)

Along with these, we include a fourth important result here.

* (D) The tangents at the extremities of any focal chord of a parabola intersect at right angles on thedirectrix. (Example -16)

Three tangents to a parabola form the triangle PQR. Prove that the circumcircle of PQR passes through the focusof the parabola.

Solution: We assume the parabola to be 2 4y ax= and three points on it to be 2 21 1 2 2( , 2 ), ( , 2 )A at at B at at and

23 3( , 2 ).C at at Tangents at A, C and B meet at P, Q and R as shown :

y

x

R

P

QX

A at , at( 2 )121

C at , at( 2 )32

3

Fig - 27

We need to show that the circumcircle of passes through the focus , or equivalently, the point is the same as .

PQRF

XF

B at , at( 2 )222

The co-ordinates of the points P, Q and R are respectively 1 3 1 2 2 3 2 3( , ( )), ( , ( )at t a t t at t a t t+ + and

1 2 1 2( , ( ))at t a t t+ .

To show that the circumcircle passes through F, it would suffice to prove that the chord PQ subtendsthe same angle on F as it does on R. Since a chord of a circle subtends equal angles anywhere on thecircumference, this will prove that F also lies on the circumference of the circle.

Example – 21



To evaluate ,PRQ∠ we need the slopes of PR and RQ which are simply the slopes of the tangents atA and B respectively, i.e.

1

1PRm

t= and

2

1RQm

t=

Thus,

tan( )1

PR RQ

PR RQ

m mPRQ

m m−

∠ =+

2 1

1 21t t

t t−=

+ ...(1)

Now, to evaluate the angle that the chord PQ subtends at F, we need to slopes PFm and QFm :

1 3

1 3

( )PF

a t tmat t a

+=−

and 2 3

2 3

( )QF

a t tmat t a

+=−

1 3

1 3 1PFt tmt t

+⇒ =−

and 2 3

2 3 1QFt tmt t

+=−

Thus,

tan( )1

PF QF

PF QF

m mPFQ

m m−

∠ =+

1 3 2 3 2 3 1 3

1 3 2 3 1 3 2 3

( )( 1) ( )( 1)( 1)( 1) ( )( )t t t t t t t tt t t t t t t t

+ − − + −=− − + + +

2 1

1 21t t

t t−=

+ ...(2)

Comparing (1) and (2) gives PRQ PFQ∠ = ∠ which confirms that F does indeed lie on thecircumference of PQR s′∆ circumcircle.

We could alternatively have done this question by explicitly evaluating the equation of the circle passingthrough P, Q and R and showing the F satisfies that equation.



Find the equation of the common tangent(s) to

(a) 2 4y ax= and 2 2 2( )x a y a+ + =

(b) 2 4y x= and 2 4x y=

Solution: (a) The equation of any tangent to 2 4y ax= can be written as

2ty x at= +

If this is to touch the circle too, the distance of the circle�s centre, i.e. (a, 0) from this line must beequal to its radius a. Thus,

2

21

a ata

t

− +=

+

2 2 2(1 ) (1 )t t− = +

4 2 21 2 1t t t⇒ + − = +

2 2( 3) 0t t⇒ − =

3t⇒ = ± (t = 0 is not a feasible solution for this case)

Thus, the required common tangents are

3 3y x a± = +

(b) Any tangent to 2 4y x= can be assumed to be of the form

2ty x t= + ( since a = 1 for this case )

If this line is to be tangent to 2 4x y= too, its intersection with this curve must yield only one point.Thus,

22

4xt x t

= +

must yield only one root, i.e. its D must equal O.

This gives

316 16 0t+ =

1t⇒ = −

Example – 22



The required common tangent is thus

1 0x y+ + =

Fig - 28

y

xThe common tangentto the two parabolas,

1 0x + y + =

y = x2 4

B

Ax = y2 4

Find the angle(s) at which the curves 2 4y x= and 2 4x y= intersect.

Solution: Recall how we defined the angle of intersection of two curves in the unit on Circles : the anglesbetween their respective tangents at the point of intersection. It is evident by simple substitution thatthe two parabolas given intersect at (0, 0) and (4, 4).

Fig - 29

y

x

(4,4)θ

(0,0)

The tangents to the two parabolas at the origin can be seen to be perpendicular by inspection. Thetwo curves therefore intersect at 90° at the origin.At (4, 4), the tangent to 2 4y x= has the slope

1(4,4) (4,4)

2 12

dymdx y

= = =

and the tangent to 2 4x y= has the slope

2(4,4) (4,4)

22

dy xmdx

= = =

Thus, the angle of intersection at (4, 4) is given by12 32tan 1 41 22

−θ = =

+ ⋅1tan (3 / 4)−⇒ θ =

Example – 23



Tangents PQ and PR are drawn to 2 4y ax= from an external point P. Another tangent is drawn at a point S onthe parabola. Perpendiculars from P, Q and R are dropped onto this tangent, and their lengths are p1, p2 and p3.Prove that 2

1 2 3p p p= .

Solution: The following diagram makes the situation clear.

y

x

R

p1

Fig - 30

The tangent drawn at is the line . Perpendiculars from

, and onto are of length , and respectively. We need to prove that

ST

P Q R Tp p p1 2 3

21

p = p p2 3

F2

S

U

F1

Q

P

F3

V

p3

p2

We assume the points Q and R to be 21 1( , 2 )at at and 2

2 2( , 2 )at at so that P is 1 2 1 2( , ( )).at t a t t+

Let S be the point t, so that the tangent at S is

2ty x at= +

We now evaluate the three perpendiculars p1, p2 and p3.

21 2 1 2 1 2

1 2 2

( ) ( )( )

1 1

at t at t t at a t t t tp

t t

− + + − −= =

+ +2 2 21 1 1

2 2 2

2 ( )1 1

at att at a t tpt t

− + −= =+ +

2 2 22 2 2

3 2 2

2 ( )1 1

at att at a t tpt t

− + −= =+ +

It is now evident that 21 2 3.p p p=

Example – 24



Referring to the figure of Example 24 above, prove that

1PU PVPQ PR

+ =

Solution: Note that

2 1PUF QUF∆ ∆∼

2 1PF QFPU QU

⇒ =

2 1

1 2

PF PU pQF QU p

⇒ = =2 1 1 2( , )PF p QF p= =∵

1

1 2

PU pPQ p p

⇒ =+

Similarly, 2 3PVF RVF∆ ∆∼ so that

1

1 3

PV pPR p p

=+

Thus,

1 1

1 2 1 3

PU PV p pPQ PR p p p p

+ = ++ +

1 1 2 3

21 1 2 1 3 2 3

(2 )p p p pp p p p p p p

+ +=+ + +

2 3 1 2 1 3

2 3 1 2 1 3

22

p p p p p pp p p p p p

+ +=+ +

21 2 3Using from

the last examplep p p =

= 1

Example – 25



Q. 1 A, B and C are three points on the parabola 2 4y ax= whose y-coordinates are in GP. Prove that thetangents at A and C will meet at a point whose y-coordinates is the same as that of point B.

Q. 2 A chord of the parabola 2 4y ax= subtends a right angle at the vertex. What is the locus of the pointof intersection of tangents at its extremities ?

Q. 3 From an arbitrary point P on the directrix, a tangent PQ is drawn to the parabola 2 4y ax= . Find thelocus of the mid-point of PQ.

Q. 4 Prove that the mid-point of the intercept made on a tangent to a parabola by the tangents at two fixedpoints A and B lies on the tangent which is parallel to AB.

Q. 5 If two tangents to a parabola intercept a constant length on any fixed tangent, prove that the locus oftheir point to intersection is another parabola with its LR being of the same length as the originalparabola.

TRY YOURSELF - II



Many properties of normals in parabolas are non-trivial and important enough to be discussed independently oftangents. This is what we do here. Once again, we use the parabola 2 4y ax= for illustration purposes althoughthe discussion can easily be generalised.

EQUATION OF : Let 1 1( , )P x y be a point on the parabola; thus 21 14y ax= .

NORMALAT P(x1, y1) The slope of the tangent at P is

1 1( , ) 1

2T

x y

dy amdx y

= =

Thus, the slope of the normal is

1

2Nyma

−=

The equation of the normal at 1 1( , )x y can now be written using point-slope form:

11 1( )

2yy y x xa

−− = −

EQUATION OF : The point P is now given in parametric form. Using the equation of the normal

NORMAL AT P(at2,2at) derived above, we can write the equation in parametric form by using (at2, 2at)instead of (x1, y1) :

222 ( )2aty at x ata

− = − −

32y tx at at⇒ + = +

Note that this normal has slope m = �t. Thus, the same equation can also bespecified in terms of slope as described below.

EQUATION OF : Instead of t, we use �m in the equation above :NORMAL WITH 32y mx am am− = − −SLOPE m 32y mx am am⇒ = − −

Note that this is the normal at the point (at 2, 2at) which in terms of m is (am2, �2am).

Section - 3 NORMALS



The cubic equation in m hints that from a given point P, three normals can be drawn to the parabola. Let us try toprove this. Suppose that the point P is (h, k). Since the normal (s) of slope m passes through P, we have

32k mh am am= − −

3 (2 ) 0am a h m k⇒ + − − = ...(1)

This gives three values of m, say m1, m2 and m3 and thus three corresponding normals. However, the roots m1, m2and m3 may not all be real. Two of them could be imaginary (one will always be real). Thus, depending on thecoefficients in(1), we could either have one or three normals from P to the parabola. (There is also the possibilityof two identical roots in which case only two normals will actually exist)

In case there are three normals, these will intersect the parabola at 2( , 2 )i iam am− for i = 1, 2 3. The sum of theordinates of these points is

1 2 32 ( )a m m m− + +

which is 0 from (1).

What are the points on the parabola 2 4y ax= from which three distinct normals can be drawn to the parabola?

Solution: Assume a point 2( , 2 )P at at from which three normal to the parabola can be drawn. We basicallyneed to find the range of the variable t for which this is possible. The equation of an arbitrary normalto this parabola can be written as

32y mx am am= − −

If this passes through P, we have

2 32 2at at m am am= − −

3 2(2 ) 2 0m t m t⇒ + − + =

2( )( 2) 0m t m mt⇒ + − + =

One root for m is �t which actually gives the normal at P itself. This should have been expectedbecause a normal to the parabola itself on any point can obviously always be drawn.

The other two roots for m are real and distinct if

t2 > 8

This is the condition that the parameter t must satisfy if we are to be able to draw three real and distinctnormals from P to the parabola.

A normal is drawn to the parabola 2 4y ax= at the point t. Find the other point at which this normal intersects theparabola.

Example – 26

Example – 27



Solution: The equation of the normal at t is

32y tx at at+ = + ...(1)

If this intersects the parabola again at 21 1 1then ( , 2 )t at at must satisfy (1). Thus,

2 31 12 2at at t at at+ = +

2 21 1( ) 2 ( )at t t a t t⇒ − = −

1( ) 2t t t⇒ + = −

12t tt

⇒ = − −

This very frequently used result tells us how to find the other end point of the normal chord which isnormal at a point t to 2 4 .y ax=

Sometimes, a question might be posed pertaining to normals. Instead of solving it entirely from scratch, we coulduse already known properties that we might have discussed earlier, say, in the section on tangents. This saves a lotof time in a subject like co-ordinate geometry.For example, suppose AB is a focal chord of a parabola. What is the angle between the tangent at A and thenormal at B?We discussed earlier that tangents at extremities of any focal chord are perpendicular. Thus, a tangent at oneend-point and a normal at the other must be parallel !Thus, you can see that the skill you need to master for this subject is to remember certain well known and

frequently used results and use them to your advantage as much as possible.

The normals at t1 and t2 of the parabola 2 4y ax= meet at t3. Prove that 1 2 2.t t =

Solution: We can very easily solve this question using the result of the last example. Since the normal drawn at

t1 intersects the parabola again in t3, we have 3 11

2 .t tt

= − − Similarly, 3 22

2 .t tt

= − − Comparing the

two gives t1t2 = 2. But let us solve it without using this result. The equation of the normals at t1 and t2are

31 1 12y t x at at+ = + ...(1)

32 2 22y t x at at+ = + ...(2)

Both of these must be satisfied by the point t3, i.e., by 23 3( , 2 ).at at Thus,

2 33 1 3 1 12 2at at t at at+ = + ...(3)

2 33 2 3 2 22 2at at t at at+ = + ...(4)

Example – 28



Subtracting(4) from (3), we have3 3 33 1 2 1 2 1 2( ) 2 ( ) ( )at t t a t t a t t− = − + −

2 2 23 1 2 1 22t t t t t⇒ = + + + ...(5)

Using (5) in (3), we finally have2 2 2 2 2 2

1 2 1 2 1 2 1 24 (2 ) ( )a t t t t at t at t+ + + = +2 2 2 2

1 2 1 2 1 2 1 28 4{( ) } ( )t t t t t t t t⇒ + + − = +2

1 2 1 2 1 2( 2){( ) ( 2) 4} 0t t t t t t⇒ − + + + =

which gives 1 2 2,t t = as required. ____________________________________________________________________________________On page - 29, four important properties that tangents in any parabola satisfy have been listed. Since a normal at apoint is perpendicular to the tangent at that point, we can use the properties of tangents to deduce the correspondingproperties for normals.* (A) Tangent at any point bisects the angle θ Normal at any point bisects the external angle

between the focal chord through that point ⇒ between the focal chord and the perpendicularand the perpendicular to the directrix from to the directrix from that point, i.e. it bisects thethat point supplementary angle 180 � θ.

* (B) The tangent at one extremity of any The tangent at one extremity of any focal chordfocal chord of a parabola is perpendicular ⇒ of a parabola is parallel to the normal at the otherto the normal at the other extremity. extremity. (We�ve already stated this earlier)

Note that from property (A) above, we can also deduce that the normal at any point of a parabola is equallyinclined to the focal chord through that point and the axis of the parabola. You are urged to prove this independentlyas an exercise.

Find the locus of the point of intersection of normals drawn to the parabola 2 4y ax= at the extremities of a chordwhich subtends a right angle at the vertex of the parabola.

Solution: Let 21 1( , 2 )A at at and 2

2 2( , 2 )B at at be the extremities of a chord which subtends a right angle at thevertex (0, 0):

Fig - 31

y

x

ABO

AB, N N C

C

is a chord whichsubtends a right angle at .The normals at and

and , intersect at . We need to find the locus of .

1 2O

N2

N1

C

21A at , at( 2 )1

22B at , at( 2 )2

Example – 29



Since ,OA OB⊥ we have

1 22 21 2

slope of slope of

2 0 2 0 10 0

OA OB

at atat at

− − × = − − − $%&%' $%&%'

1 2 4t t⇒ = −

The equation to N1 and N2 can be written using the standard form of a normal at a point t :

31 1 1 1: 2N y t x at at+ = +

32 2 2 2: 2N y t x at at+ = +

Let the intersection of N1 and N2 be the point C(h, k). The co-ordinates of C can be evaluated bysolving the equations of N1 and N2 simultaneously :

2 21 2 1 22 ( )h a a t t t t= + + +

2 21 22 ( )a a t t= − + +

and 1 2 1 2( )k at t t t= − +

1 24 ( )a t t= +

We thus have, by eliminating t1 and t2, a relation in h and k:

2

2

2 816

k h aa a

+= −

2 16 ( 6 )k a h a⇒ = −

Using (x, y) instead of (h, k), the required locus is

2 16 ( 6 )y a x a= −

Find the locus of the point of intersection of the three normals to the parabola 2 4 ,y ax= two of which are inclinedat right angles to each other.

Solution: Let P(h, k) be the point whose locus we wish to determine. Any normal to the given parabola can bewritten as


If this passes through P(h, k), we have

32k mh am am= − − ...(1)

Example – 30



Let m1, m2 and m3 be the three roots of this cubic. It is given that two of the normals are perpendicular,implying that the product of two of these three slopes, say m1 and m2 is �1, i.e. m1m2 = �1.

From (1), we have

1 2 3km m m

a−=

3kma

⇒ =

Substituting this value of m3 back in (1), we obtain a relation between h and k :

3

32hk akk ka a

= − −

2( (3 ) ) 0k k a h a⇒ + − =

Using (x, y) instead of (h, k) the required locus is

2( (3 ) ) 0y y a x a+ − =



Q. 1 Find the equation of the normal to 2 4y x= which is perpendicular to 2 6 5 0x y+ + = .

Q. 2 The normal at any point P on the parabola 2 4y ax= meets its axis in A and the y-axis in B. Let O bethe origin. The rectangle OACB is completed. Find the locus of C.

Q. 3 Prove that the normal to 2 4y ax= at a non-zero point whose x and y coordinates are equal, subtendsa right angle at the focus.

Q. 4 Prove that the normal at any point to a parabola is equally inclined to the focal chord passing throughthat point and the axis of the parabola.

Q. 5 Have you ever heard of parabolic mirrors being very effective in concentrating incident light energy ata particular a point ? Can you think of a reason ? What would that particular point be ?

TRY YOURSELF - III



In the unit on circles, we derived some significant results:

� Pair of tangents from (x1, y1) : 21 1 1 1( , ) ( , ) ( , )T x y S x y S x y=

� Chord of contact of the two : 1 1( , ) 0T x y =tangents drawn from (x1, y1)

� Chord bisected at a given : 1 1 1 1( , ) ( , )T x y S x y=point (x1, y1)

In this section we�ll see that these results can be generalised to the parabola in the same forms !

Result - 1 : We first of all prove that two real and distinct tangents can be drawn from any externalpoint to a given parabola. Let the parabola 2 4y ax= and P(h, k) be an external point

so that 2 4 .k ah>

The equation of any tangent to the parabola can be written as

ay mxm

= +

If this passes through (h, k), we have

ak mhm

= +

2 0hm km a⇒ − + =

This quadratic obviously has two real and distinct values of m as roots since thediscriminant, 2 4 ,k ah− is positive.

This analysis also confirms the trivial result that from any point inside a parabola, notangent can be drawn onto it.

Result - 2 : We wish to determine the joint equation of the pair of tangents drawn from anPair of tangents external point P (h, k) to the parabola 2 4 .y ax=

The equation of any tangent to 2 4y ax= can be written as ay mxm

= +

If this passes through P(h, k), we must have

ak mhm

= +

2 0m h km a⇒ − + =

Section - 4 MISC. RESULTS

from P (h, k)



Let m1 and m2 be the two roots of this quadratic. Thus,

1 2 1 2,k am m m mh h

+ = = ...(1)

The joint equation can now be written as

1 21 2

0a ay m x y m xm m

− − − − =

2

2 2 1 21 2 1 2

2 1 1 2 1 2

( ) 0m m a a ay m m x m m xy a x ym m m m m m

+ − + + + − + + =

All the coefficients can be expressed in terms of 1 2( )m m+ and 1 2( ).m m Thus, using(1), we obtain

2 22 2 0ax kxy k ah hy x ky

h h h a −+ − + − + =

2( ) ( )( )a x h y k xk yh⇒ − = − −

This can be written as

2 2 2( 4 )( 4 ) { 2 ( )}y ax k ah yk a x h− − = − + .... (2)

Now, if we denote 2 4y ax− as ( , )S x y and 2 ( )yk a x h− + as ( , ),T h k this relationwe obtained in (2) can be written concisely as

2( , ) ( , ) ( ( , ))S x y S h k T h k=

which is the same form as that of the circle. Thus, we can write the pair of tangents from(h, k) as

2( ( , )) ( , ) ( , )T h k S x y S h k=

which is sometimes more simply written as 21.T SS=

Chord of contact : To evaluate the chord of contact from P(h, k) to the parabola 2 4 ,y ax= we�ll followfrom P(h, k): the same approach as we did for circles. Let the tangents from P to

the parabola touch it at 1 1( , )A x y and 2 2( , ).B x y We can, using the equationof tangent at a point, write the equations of both PA and PB as

1 12 ( )yy a x x= +

2 22 ( )yy a x x= +



Since both these lines pass through P(h, k), we have

1 12 ( )ky a h x= + ...(1)

2 22 ( )ky a h x= + ...(2)

Now comes the crucial step; from (1) and (2), we can conclude that both 1 1( , )A x yand 2 2( , )B x y satisfy the linear equation

2 ( )ky a h x= +

which must then be what we are looking for: the chord of contact.

As in the case of circles, the equation to the chord of contact can be written concisely as

( , ) 0T h k =

Chord bisected : Any line passing through (h, k) can be written asat (h, k) ( )y k m x h− = − ...(1)

The intersection of this line with the parabola will, in general, give two points, say

1 1( , )A x y and 2 2( , ).B x y The co-ordinates of A and B can be obtained by

simultaneously solving the equation of this line with that of the parabola, i.e. 2 4 .y ax=Thus

2( ( ) ) 4m x h k ax− + =

2 2(2 ( ) 4 ) ( ) 0m x m k mh a x k mh⇒ + − − + − =

The roots of this equation will be x1 and x2 so that

1 2 2

4 2 ( )a m k mhx xm

− −+ =

This must equal 2h, since h is the mid-point of A and B. Thus,

2

2

4 2 22 a mk m hhm

− +=

2amk

⇒ =

Substituting this in (1), we get the required equation of the chord as 2 ( )ay k x hk

− = −

2 2 ( )yk k a x h⇒ − = −



We write this as

22 ( ) 4yk a x h k ah− + = −

because now this equation has the concise representation

( , ) ( , )T h k S h k= ,

the same as in the case of circles.

Find the locus of the mid-points of the normal chords of the parabola 2 4y ax= .

Solution: Let P(h, k) be the point whose locus we wish to determine. If a chord passing through P is bisectedat P, then its equation is, as discussed earlier,

( , ) ( , )T h k S h k=

22 ( ) 4yk a x h k ah⇒ − + = −

22 2ky ax k ah⇒ − = − ...(1)

Since this chord is also a normal to the same parabola, its equation can also be written as

32y mx am am− = − − ... (2)

(1) and (2) represent the same chord in two different forms, so that

2

3

2 21 2k a k ah

m am am−= =

− −

m can easily be eliminated using these relations to obtain a relation in h and k as

2 2 2 2 2( 2 ) 4 ( 2 ) 0k k ah a k a− + + =

Thus, the required locus is

2 2 2 2 2( 2 ) 4 ( 2 ) 0y y ax a y a− + + =

From a point A common tangents are drawn to the circle 2

2 2

2ax y+ = and the parabola 2 4 .y ax= Find the area

of the quadrilateral formed by the common tangents, the chord of contact of the circle and the chord of contact ofthe parabola.

Example – 31

Example – 32



Solution: The figure below makes the situation described in the question clearer :

Fig - 32

y

The chord of contact of the circle is while that of the parabola is .

We need to find the area of the quadrilateral

PQ RS

PQSR.

R

P

AQ

S

x

y = ax2 4

Any tangent to 2 4y ax= can be written in terms of its slope m as

ay mxm

= +

or 2 0m x my a− + =

If this is to touch the circle 2

2 2 ,2ax y+ = its distance from the circle�s centre (0, 0) must be equal to

the circle�s radius .2

a

Thus,

4 2 2a a

m m=

+4 2 2m m⇒ + =

2 2( 2)( 1) 0m m⇒ + − =

1m⇒ = ±

Thus, the two tangents from A are ( )y x a= ± + so that the point A is (�a, 0).

The chord of contact PQ is

( ,0) 0CT a− =

2

0 aax yy

⇒ − + =

2ax⇒ = −



The chord of contact RS is

( ,0) 0PT a− =

0 2 ( )y a x a⇒ = −

x a⇒ =

The length PQ is now simply a (verify) whereas RS becomes the Latus-rectum (since x = a for RS) so

that its length is 4a. The distance between PQ and RS is 3 .

2 2a aa+ =

Thus, the area of PQSR (note that it is a trapezium) is

1 3( 4 )2 2

aa a∆ = + ×

215 sq. units

4a=

Find the locus of an external point P from which exactly two distinct normals can be drawn to the parabola2 4 .y ax=

Solution: The equation of the normal in terms of slope is a cubic, from which we inferred that in general threenormals can be drawn to the parabola from a given point. Thus, we could either have one real and twoimaginary solutions of the cubic (implying only one normal) or all the three roots of the cubic could bereal (implying three real normals).

In this question, we have exactly two distinct normals which can only happen if all the three roots ofthe cubic are real and two of these three roots are identical. This is the insight we use for solvingthis question.

The equation of an arbitrary normal to the parabola is


If this passes through ( , ),P h k we have


3 (2 ) 0am a h m k⇒ + − + = ...(1)

Let the roots of this cubic be m1, m1 and m2 so that

1 2 2 12 0 2m m m m+ = ⇒ = −

21 2

km ma

= −

Example – 33



These two relations give133

1 12 2k km ma a

= ⇒ = We now substitute this value of m1 back in (1) to obtain a relation between h and k :

13

(2 ) 02 2ak ka h ka a

+ − + =

2 327 4( 2 )ak h a⇒ = −

Using (x, y) instead of (h, k) gives the required locus as

2 327 4( 2 )ay x a= −

Find the locus of the point of intersection of tangents drawn at the extremities of a normal chord to the parabola2 4 .y ax=

Solution: Let t1 be a point at which a normal is drawn to the parabola. This intersects the parabola again at thepoint

2 11

2t tt

= − − ...(1)

Let the point of intersection of the tangents at t1 and t2 be ( , ).P h k From an earlier obtained result, wehave

1 2h at t= 1 2( )k a t t= +

1 11 1

2 2h at t k at t

−⇒ = − + =

(Using (1))

21

1

22 ah at a kt

⇒ = − − = −

2

2

2 4h a aa k

+⇒ =−

(eliminating t1)

2 3( 2 ) 4 0k h a a⇒ + + =

The required locus of P is

2 3( 2 ) 4 0y x a a+ + =

Example – 34



Find the polar of the pole 1 1( , )P x y with respect to the parabola y2 = 4ax.

Solution: Recall that the polar is the locus of the point of intersection of the two tangents drawn at the extremitiesof a variable chord QR passing through P.

Let M(h, k) be a point on the required polar corresponding to QR.

The chord of contact of the tangents drawn from M, i.e. QR, has the equation

( , ) 0T h k =

2 ( )ky a x h= +

If this is to pass through 1 1( , ),P x y we have

1 12 ( )ky a x h= +

This is the required locus; in terms of (x, y), the locus (polar) is

1 12 ( )yy a x x= +

1 1 1 1( , ) 0 : Equation of polar of ( , )T x y P x y⇒ =

Observe that the polar of the focus will simply be the directrix.

Sometimes, the equation of the parabola might not be in the standard form 2 4y ax= or 2 4 .x ay= However, bya transformation (translation and / or rotation) of the co-ordinate axes, the equation given can always be reducedto the standard form.

For example, suppose that the equation of a (translated) parabola is2( ) 4 ( )y a x− α = −β

The vertex of this parabola is at ( , )β α instead of (0, 0). Thus, we cannot use all the formulae that we�ve deriveddirectly on this form. However, by a translation of the axis, so that the new origin coincides with ( , ),β α i.e., by thetransformation

,y y x x′ ′→ + α → +β

the equation of the parabola referred to the new axes will become2 4y ax′ ′=

which is indeed in the standard form. We can now use this form to evaluate whatever we were required to, say,some tangent or some normal in the x y′ ′− axes. Once we have the required equation, we can convert it to thex � y system using the reverse transformations

,y y x x′ ′→ − α → −β

Example – 35

CONCLUDING NOTE



Fig - 33

y

x

The equation of is

( � ) = 4 ( � ) in the original system. But in the translated system ' � ', the equation of is in its standard form

' = 4 '

C

y a x

x yC

y ax

α β2

2

(0, 0)

y'

x'

C

( , )αβ

As a very simple example, suppose that we have to find the focus and directrix of the parabola

2 4 6 25 0y x y− − + =

which can be rearranged as

2( 3) 4( 4)y x− = − ...(1)

Thus, a = 1. However, we cannot directly use (a, 0) as the focus which should be obvious.

Referred to the co-ordinate system x y′ ′− where 4x x′ = − and 3,y y′ = − the equation (1) is

2 4y x′ ′=

It is in this system that the focus will be at (a, 0), i.e. at (1, 0). Thus, 1fx′ = and 0fy′ = . In the original system, thefocus can be evaluated by the reverse transformation

4x x′= + and 3y y′= +

4f fx x′⇒ = + and 3f fy y′= +

= 5 = 3

Thus, the focus in the original system is at (5, 3). Of course, this example is trivial and the answer could have beenevaluated simply be inspection. However, it was described just to highlight the underlying approach you couldfollow in such cases.

Let us now consider a non-trivial example using this approach. Suppose that we have to find all the points on theaxis of the parabola 23 4 6 8 0y y x+ − + = from where three distinct normals can be drawn to the parabola.Assume such points to be (h, k). The equation of the parabola can be rearranged as

22 1023 9

y x + = −



which can be written in the standard form using the transformations10 2,9 3

X x Y y= − = +

as2 2Y X=

Thus,12

a =

We will now work in the X-Y system. Any point on the axis of the parabola can be taken as (p, 0) (in this newsystem). The equation of an arbitrary normal to the parabola 2 2Y X= is

3

2mY mX m= − −

If this passes through (p, 0), we have

3

02

mmp m− − =

2

1 02

mm p

⇒ + − =

0, 2( 1)m p⇒ = ± −

Thus, we have three distinct normals if

1 0p − >

1p⇒ >

Therefore, three distinct normals can be drawn to the parabola from ( ,0)p if p > 1.

However, note that this result is in the X � Y system and not the x � y system. To obtain the result in the originalsystem, we must use the reverse transformations

10 2,9 3

X x Y y= − = +

Thus, if as assumed earlier, the required points in the original system are (h, k) (which are (p,0) in the new system)we have

10 209 3

p h k= − = +

Since p > 1, we have 19 .9

h > Also, 23

k = − .

Thus, from all points on the axis, 2,3

h − , where 19

9h > , three distinct normals can be drawn to the parabola.

If you did not understand this discussion, you are urged to re-read it until you do.



Q. 1 Find the locus of the mid-points of the chords of the parabola 2 4y ax= which subtend a right angleat the focus (using a homogenizing approach.)

Q. 2 Prove that the locus of the mid-point of a system of parallel chords of a parabola is a line parallel to theaxis of the parabola.

Q. 3 What is the equation of the chord of contact of the tangents drawn from ( 1, 2)P − to 2 4 ?y x=

Q. 4 Prove the pole of an external point P is the chord of contact of the tangents drawn from P to theparabola.

Q. 5 From the pole P of a chord AB of a parabola, a line is drawn parallel to the axis of the parabola. Thisline intersects the parabola in Q and AB in R. Prove that PQ = QR.

TRY YOURSELF - IV



Find the values that a can take so that the point ( 2 , 1)P a a− + lies inside the smaller region bounded by the circle2 2 4x y+ = and the parabola 2 4y x= as shown:

Fig - 34

y

xFor what values of will (� 2 , + 1)

lie inside the shadedregion ?

a a a

Solution: Note that the shaded region represents the interior of both the circle and the parabola. Thus,

2 2

so that lies in theinterior of the circle

( 2 ) ( 1) 4 0P

a a− + + − <$%%%%&%%%%' and 2

so that lies in theinterior of the parabola

( 1) 4( 2 ) 0P

a a+ − − <$%%%&%%%'

25 2 3 0a a⇒ + − < and 2 10 1 0a a+ + <

315

a⇒ − < < and 5 2 6 5 2 6a− − < < − +

The intersection of the two constraints gives the possible values of a as

( 1, 5 2 6)a ∈ − − +

Prove that the area of the triangle formed by three points on a parabola is twice the area of the triangle formed bythe tangents at these points.

SOLVED EXAMPLES

Example – 1

Example – 2



Solution:

Fig - 35

y

x We need to prove thatarea ( ) = 2 area ( )∆ABC PQR∆

P

R

Q

C

B

A

Let the parabola be 2 4y ax= .

Assume the points A, B and C to be t1, t2 and t3 respectively. Thus, P, Q and R are respectively

1 2 1 2 2 3 2 3( , ( )), ( , ( ))at t a t t at t a t t+ + and 3 1 3 1( , ( )).at t a t t+ Now we simply use the determinant formulato find the area of the two triangles

area

2 2 21 2 3

1 2 31( ) 2 2 22

1 1 1

at at atABC at at at∆ =

2 2 2 2 21 2 1 3 1

21 2 1 3 1

1 0 0

t t t t ta t t t t t

− −= − − 2 2 1

3 3 1

Using ;C C CC C C

→ − → −

21 2 1 3 1

22 1 3 1 1( )( ) 1 1

1 0 0

t t t t ta t t t t t

+ += − −

21 2 2 3 3 1( )( )( )a t t t t t t= − − − −



Similarly,

area 1 2 2 3 3 1

1 2 2 3 3 11( ) ( ) ( ) ( )2

1 1 1

at t at t at tPQR a t t a t t a t t∆ = + + +

1 2 2 3 1 1 3 22

1 2 3 1 3 2

( ) ( )( )

21 0 0

t t t t t t t ta t t t t t t

− −= + − −

1 2 2 12

3 2 3 1 1 2( )( ) ( ) 1 12

1 0 0

t t t ta t t t t t t= − − +

2

2 1 2 3 3 1( )( )( )2a t t t t t t= − − −

Clearly, area ( ) 2 area ( )ABC PQR∆ = ∆

This example should show you that the solution would have been much lengthier had we notremembered the formula for the co-ordinates of the intersection point of two tangents. This is onesubject where some things are to be learnt!

Consider a fixed parabola 2 4 .y ax= A variable but equal parabola having its axis parallel to the axis of the fixedparabola, moves so that it always touches the fixed parabola. What is the locus of the vertex of the movingparabola ?

Solution: By saying that two parabolas are equal, we can infer that the length of their latus-rectums will thesame. It should be obvious that for an equal parabola to touch 2 4 ,y ax= it must open towards the left(the axis of the variable parabola is parallel to the x-axis.):

Fig - 36

y

x We nee to find the locus of ( ) V h, k

P , ( )βα

V ( , )h k

y = ax2 4

Let the point of contact of the two parabolas be ( , ).P α β Note that at P the two parabolas will havea common tangent.

Example – 3



We have assumed the vertex of the variable parabola to be V (h, k). Thus the equation of this variableparabola is

2( ) 4 ( )y k a x h− = − − ...(1)

Now, the tangent at ( , )P α β to 2 4y ax= will be

2 ( )y a xβ = + α ...(2)

The tangent at ( , )P α β to the variable parabola will be (using (1)),

( )( ) 2 ( )y k k a x h h− β − = − − + α − ...(3)

In writing (3), we have implicitly used a translation of the axis to (h, k) ; verify that (3) is indeedcorrect. (3) can be written in standard form as

2 ( ) 2 ( 2 ) ( ) 0ax k y a h k k+ β − + α − − β − = ...(4)

Now note that (2) and (4) are in fact the same line, so that

2 22 2 ( 2 ) ( )

a aa k a h k k

−β α= =β− α − − β−

α and β can easily be eliminated from these relations to obtain a relation between h and k as

2 8k ah=

Thus, the required locus of V is

2 8y ax=

which is a parabola with the same vertex as the original parabola, and the same orientation, as onemight have intuitively expected.

Find the condition for all the three normals drawn from a given point P(h, k) to the parabola 2 4y ax= to be realand distinct. Assume a > 0.

Solution: As discussed a lot of times earlier, the given requirement is equivalent to the equation


having three real and distinct roots

Example – 4



A cubic will always have at least one real root. Here are some configurations that the graph of a cubiccan take (assuming the coefficient of m3 is positive):

Fig - 37

Some configurations that the graph of a cubic can take

Since we want three real roots, we focus on the last graph above. Observe that it has two extremas,say at α and β .

α

β

f x( )

A cubic ( ) with threereal zeroes

f x

Fig - 38

Thus, ( ) 0f x′ = must have two distinct real roots (since ( ) ( ) 0).f f′ ′α = β = Also, we must have( ) ( ) 0.f fα β < Convince yourself that these two constraints will ensure that the cubic f(x) will have

three real and distinct zeroes.

We use this for the current cubic 3( ) (2 )f m am a h m k= + − +



� ( ) 0f m′ = has two real and distinct roots

2( ) 3 (2 )f m am a h′⇒ = + −

( ) 0f m′⇒ = for 2

3h am

a−= ±

2 0h a⇒ − > or 2h a>

The two real and distinct roots are

23

h aa

−α = and 2

3h a

a−β = − ...(1)

� ( ) ( ) 0f fα β <

3 3( (2 ) )( (2 ) ) 0a a h k a h k⇒ α + − α + α β + − β+ <

2 3 2 2 3 3 2( ) (2 ) ( ) ( ) 0a a a h ak k⇒ α β + − αβ α + β + α +β + <

Simplifying this using (1), we obtain

2 327 4( 2 )ak h a< −

Thus, if we are to be able to draw three real and distinct normals form P(h, k) to 2 4 ,y ax= h and kmust satisfy

2h a> and 2 327 4( 2 )ak h a< −

From a variable point P on a fixed normal to the parabola 2 4 ,y ax= two more normals are drawn to the parabolato intersect it at Q and R. Show that the variable chord QR will have a fixed slope.

Solution: Let P be the point (h, k). An arbitrary normal to the given parabola can be written as


and if this passes through P, we have

32k mh am am= − − ...(1)

Now, this cubic has three roots, say 1 2,m m and 3,m of which one say m1, is fixed, equal to the slopeof the fixed normal. The other two slopes are variable and thus the points of intersection of the othertwo normals through 2

2 2{ ( , 2 )P Q am am− and 23 3( , 2 )}R am am− are also variable.

Example – 5



From (1), we have

1 2 3 0m m m+ + = ...(2)The slope of QR is

2 32 23 2

2 ( )( )a m mm

a m m−=

−

2 3

2( )m m

=− +

1

2m

= (from (2))

This shows that m, the slope of QR, is fixed since m1 is fixed.

Find the locus of the point of intersection of those normals to the parabola 2 8x y= which are at right angles toeach other.

Solution: We have already discussed a very similar question earlier. The point of including this question here isthat the equation to the parabola has been given not in the standard form 2 4y ax= but in the form

2 4 :x ay= this means that in all the formulae that we use, x and y must be interchanged. Thus, the

equation of an arbitrary normal to 2 8x y= can be written as

34 2x my m m= − −

If this passes through ( , ),P h k we obtain

34 2h km m m= − − ...(1)

This cubic has three roots, say 1 2,m m and 3.m Since two of the normals are perpendicular, we can

take 1 2m m equal to �1.

From (1), we have

1 2 3 0m m m+ + =

1 2 2 3 3 1 22km m m m m m+ + = − +

1 2 3 2hm m m = −

Using 1 2 1m m = − and these three equations, a relation involving only h and k can easily be obtained:2 2 12 0h k− + =

Thus, the required locus is2 2 12 0x y− + =

Example – 6



The normals at three points A, B and C on a parabola intersect at O. F is the focus of the parabola. Prove that

2

4lFA FB FC FO⋅ ⋅ = ⋅

where l is the length of the latus-rectum.

Solution: Assume the parabola to be 2 4 ,y ax= and the point O to be (h, k). F is the point (a, 0).

Fig - 39

y

x

A

O h, k( )

B

F

C

Any normal to 2 4y ax= is


and since this passes through O(h, k), we have

32k mh am am= − − ...(1)

The feet of the three normal correspond to A, B and C. Thus, if 1 2,m m and 3m are the roots of (1),the co-ordinates of A, B and C are

2 21 1 2 2( , 2 ), ( , 2 )am am am am− − and 2

3 3( , 2 )am am−

We have now,

2 2 21 1( ) (2 )FA a am am= − +

21(1 )a m= +

Similarly, 22(1 )FB a m= + and 2

3(1 )FC a m= + .

Example – 7



Thus,

3 2 2 21 2 3(1 )(1 )(1 )FA FB FC a m m m⋅ ⋅ = + + +

( )3 2 2 2 2 2 2 2 2 2 21 2 3 1 2 2 3 3 1 1 2 31 ( )a m m m m m m m m m m m m= + + + + + + + ...(2)

From (1), we know the values of

1 2 3

1 2 2 3 3 1

1 2 3

02

m m ma hm m m m m ma

km m ma

+ + =

− + + =

= −

...(3)

Our task is to express the relation (2) in terms of the known quantities given by (3). This can be doneas follows :

2 2 3 21 2 3 1 2 3 1 2 2 3 3 1( ) 2( )m m m m m m m m m m m m+ + = + + − + +

2 2 2 2 2 2 21 2 2 3 3 1 1 2 2 3 3 1 1 2 3 1 2 3( ) 2 ( )m m m m m m m m m m m m m m m m m m+ + = + + − + +

Substituting the appropriate values gives

2 23

2

2 21 2a h a h kFA FB FC aa a a

− − ⋅ ⋅ = + − +

2 2{( ) }a h a k= − +

which evidently equals 2.a FO⋅ Since 4 ,l a= we get the desired result:

2

4lFA FB FC FO⋅ ⋅ = ⋅

Normals are drawn form the point P with slopes m1, m2, m3 to the parabola 2 4 .y x= If the locus of P with

1 2m m = α is a part of the parabola itself, then find α .

Example – 8



Solution: If we let P be the point (h, k), we have

3 (2 ) 0m h m k+ − + = ...(1)

Thus,

1 2 3m m m k= −

3km⇒ = −α (since 1 2m m = α )

Substituting this back into (1), we obtain

3

3 (2 ) 0k k h k− − − + =α α

2 2 2 32k h⇒ = α − α + α ...(2)

Also, since P lies on the parabola itself, we have

2 4k h= ...(3)

From (2) and (3), we have

2 4α = and 3 22 0α − α =

2⇒ α =

Let AB be a fixed chord passing through the focus of a parabola. Prove that three circles can be drawn whichtouch the parabola and AB at the focus.

Solution: Let the parabola be 2 4 ;y ax= its focus is then F(a, 0). The following diagram shows one such circlewhich touches AB at F and also the parabola :

Fig - 40

y

x

A

y ax2 = 4

O F

BP

Example – 9



Let AB have a (fixed) slope m. The equation of AB can be written using the point-slope form :

0 ( )y m x a− = −

y mx am⇒ = −

Any circle touching this line at F(a, 0) can be written in terms of a real variable λ as :

2 2( ) ( ) 0x a y y mx am− + + λ − + =

2 2 (2 ) ( ) 0x y a m x y a a m⇒ + − + λ + λ + + λ = ...(1)

We want those values of λ for which this circle touches the parabola.

Assume the point of contact of the circle and the parabola to be 2( , 2 ).P at at Thus, at P, the circle andthe parabola should have a common tangent.

The tangent at P to the parabola is

2ty x at= +

Since this line is a tangent to the circle as well (at the same point 2( , 2 )),P at at we can write theequation of the same circle using another real parameter :α

2 2 2 2( ) ( 2 ) ( ) 0x at y at ty x at− + − + α − − =

2 2 2 2 4 2 2 2( 2 ) ( 2 ) 4 0x y at x t at y a t a t a t⇒ + − α + + α − + + − α = ...(2)

The circles given by (1) and (2) being the same, we can compare the coefficients to obtain the followingequations :

22 2a m at+ λ = α +

2t atλ = α −

4 2 24a m at at t+ λ = + − α

We can rearrange these equations to make them look more �systematic�.

2( ) ( 1) (2 2 ) 0m a atλ + − α + − =

(1) ( ) (2 ) 0t atλ + − α + =

2 2 4( ) ( ) ( 4 ) 0m t a at atλ + α + − − =

The variables λ and α can now be eliminated to obtain a relation purely in terms of t:

2

2 4 2

1 2 (1 )1 2 0

( 4 1)

m a tt at

m t a t t

− −− =

− + −



Expanding along C3 and simplifying, we obtain

4 2 2 4( 2 3) 1 2 3 0mt t t t t− − + − − =

2 2 2 2( 1)( 3) (3 1)( 1)mt t t t t⇒ + − = − +

2 2( 3) 3 1mt t t⇒ − = −

3 23 3 1 0mt t mt⇒ − − + = ... (3)

This is a cubic in t which will have in general three roots. This will imply that three possible points ofcontact 2( , 2 ),at at and therefore three possible circles exist satisfying the given property.

But something is missing ! We still have to prove that the cubic will actually yield three real values oft. For that, we follow the approach described in Example - 4

Let 3 2( ) 3 3 1f t mt t mt= − − +

First, we show that ( )f t′ has two real roots, say t1 and t2.

2( ) 3 6 3f t mt t m′ = − −

2( ) 0 2 0f t mt t m′ = ⇒ − − =

24 4 0D m⇒ = + >

Thus, ( )f t′ has two real and distinct root t1 and t2.

1 2 1 22 , 1t t t tm

⇒ + = = − ...(4)

Now,

3 2 3 21 2 1 1 1 2 2 2( ) ( ) ( 3 3 1)( 3 3 1)f t f t mt t mt mt t mt= − − + − − +

which upon simplification yields (using (4))

21 2 2

4( ) ( ) 4 2f t f t mm

= − − −

which is evidently always negative.

Thus, the cubic (3) will always give three real values of t, and hence three corresponding circles.

You may rest assured that you�ll not encounter a question with this much involved analysis in anyexam! This was included here for illustration only.



Let C1 and C2 be the parabolas 2 1x y= − and 2 1y x= − respectively. Let P be any point on C1 and Q be anypoint on C2. Let P1 and Q1 be the reflections of P and Q respectively in the line y = x. Prove that

1 1min{ , }.PQ PP QQ≥ Also, let the points P0 and Q0 on C1 and C2 respectively be such that 0 0P Q PQ≤ for allpairs of points (P, Q) with P on C1 and Q on C2. Find P0Q0.

Solution: The symmetrical nature of the situation tells us that P1 will lie on C2 and Q1 will lie on C1:

Fig - 41

y

x

C :x = y – 12 1

P1

Q

Q1

P

O

C :y = x – 22 1

X

Y

Observe from the figure thatPO PX≥

and QO QY≥

so that PO QO PQ PX QY+ = ≥ +

2 2 2PQ PX QY⇒ ≥ +

1 12PQ PP QQ⇒ ≥ +

Since PQ is greater than the mean of PP1 and QQ1, it must be greater than (or equal) to the lesser ofthe two.

To find the minimum possible value of PQ, i.e. P0Q0, observe that this length will be minimum whenPQ becomes precisely perpendicular to y = x and PQ is normal to both the parabolas at the points ofcontact P and Q. In such a configuration, assume the co-ordinates of P to be 2( , 1)t t + so that Q

(which is P's mirror image in such a configuration) will be 2( 1, ).t t+ The slope of PQ is this case

is �1. Also, the tangent to 1C at P will have slope 2 .p

dy tdx

= Thus,

2 1 1t ×− = −

12

t⇒ =

Example – 10



P and Q therefore have the co-ordinate 1 5,2 4

and 5 1,4 2

respectively. The minimum length of PQ

is now simply obtained using the distance formula :

2 2

min1 5 5 12 4 4 2o oP Q PQ = = − + −

32 2

=



ANSWERS

TRY YOURSELF -I

1. 2 216 8 96 554 1879 0x y xy x y + + + − − =

2. 2 2 14 2 2 17 0x y x y xy + − + + + =

3. 2 2y ax =

4. 8 3a

5.28 4 2

9 9 9y x

− = −

6. 2 2 ( )y a x a = −

7. ( )( )2 2 2 2 24 4y a ax y a c + − =

9. 2 28 4 0y x ax + + =

10. ( )2 4 8y a x a = −

TRY YOURSELF -II

2. [ ]4 0x a+ =

3. ( ) ( )22 2 3y x a a x a + = +

TRY YOURSELF -III

1. [ ]3 33y x= −

2. 3 2 22x ax ay = +

TRY YOURSELF -IV

1. 2 22 8 0y ax a − + =

3. [ ]1y x= −



ASSIGNMENTASSIGNMENTASSIGNMENTASSIGNMENTASSIGNMENT

[ LEVEL - I ]

1. Find the vertex, axis, focus and directrix of the parabola 2 8 19 0.y y x− − + =

2. Let 1 2,y y and 3y be the y-coordinates of the vertices of a triangle inscribed in 2 4 .y ax= Show thatthe area of the triangle is

1 2 2 3 3 11 ( )( )( )

8y y y y y y

a∆ = − − −

3. Find the focus of the parabola whose parametric equations are

( )cosx u t= α 21( sin )2

y u t gt= α −

4. Let 2( , 2)a a − be a point inside the region bounded by 2 2y x= and the line 4.x y+ = Find thepossible values of a.

5. Two perpendicular chords are drawn from the origin O to the parabola 2 ,y x x= − which meet theparabola at P and Q. The rectangle POQR is completed. Find the locus of the vertex R.

6. If PQ and RS are two focal chords of 2 4 ,y ax= prove that PR and QS intersect on the directrix ofthis parabola.

7. Find the equations of the common tangents to the parabola 2 16y x= and the circle 2 2 8.x y+ =

8. The tangent at a point P to 2 4y ax= meets the axis of the parabola at T and the normal at P cuts thecurve again in Q. If the x-coordinates P is 4a, prove that : 4 : 5PT PQ =

[ LEVEL - II ]9. Prove that if two straight lines, one a tangent to 2 4 ( )y a x a= + and the other to 2 4 ( )y b x b= + are

at right angles to one another, then they will meet on the line 0.x a b+ + =

10. A variable line passing through a fixed point A cuts a given parabola at P and Q. A variable point Rmoves on the line APQ such that AP, AR and AQ are in H.P. Prove that the locus of R is a straight line.

11. Prove that the shortest normal chord of 2 4y ax= has a slope 1tan 2− and is of length 6 3 .a



12. From a fixed point P, variable chords are drawn to meet a fixed parabola at points A and B. A circledrawn with AB as diameter meets the parabola again in C and D. Show that the line joining C and Dwill always pass through a fixed point.

13. The sides of a triangle touch the parabola 2 4y ax= and two of its angular points lie on the parabola2 4 ( ).y b x c= + Find the locus of the third angular point.

14. The normals to 2 4y ax= at Q and R meet the parabola again at P. If T is the intersection point of thetangents to the parabola at Q and R, then show that the locus of the centroid of TQR∆ is

2 (3 2 ).y a x a= +

15. From any point P, three normals are drawn to a given parabola. Show that the sum of the angles madeby these normals with the axis of the parabola exceeds the angle made by PF (F is the focus) with theaxis of the parabola by an integral multiple of .π

16. Through the vertex A of 2 4 ,y ax= two chords AP and AQ are drawn, and the circles on AP and AQ

as diameters intersect in R. Prove that if 1 2,θ θ and φ be the angles made with the axis of the parabolaby the tangents at P, Q and by the line AR, then

1 2cot cot 2 tan 0θ + θ + φ=

17. A focal chord of the parabola 2 4 ,y ax= whose perpendicular distance from the origin is b is of lengthc. Show that 2 34 .cb a=

18. Let P be any point on 2 4y ax= between its vertex and the extremity of its latus rectum lying abovethe x-axis. Q is the foot of the perpendicular from the focus F to the tangent at P. Find the maximumvalue of area ( ).FPQ∆

19. Prove that the parabolas 2 4y ax= and 2 4 ( )y c x b= − will have a common normal only if 2.ba c

>−

20. The angle between the tangents drawn from a point P to y2 = 4x is 4π

. Find the locus of P.



ASSIGNMENT

ANSWERS

LEVEL - I

1.13 11(3, 4), 4, , 4 ,4 4

y x = =

3.2 2 2sin cos 2,

2 2u u

g g α − α

4. ( 2, 2 2) − − +

5. 2 3 4y x x = − +

7. [ ]( 4)y x= ± +

LEVEL - II

13.2 24 4(2 ) bcy b a x

a a = − +

18. 2a

20. 2 2 26 0x y ax a− + + =

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